算法模板——线段树5（区间开根+区间求和）

实现功能——1：区间开根；2：区间求和（此模板以BZOJ3038为例）

作为一个非常规的线段树操作，其tag也比较特殊呵呵哒

var
   i,j,k,l,m,n:longint;
   a,b:array[0..500000] of int64;
function max(x,y:longint):longint;inline;
         begin
              if x>y then max:=x else max:=y;
         end;
function min(x,y:longint):longint;inline;
         begin
              if x<y then min:=x else min:=y;
         end;
procedure built(z,x,y:longint);inline;
          begin
               if x=y then
                  begin
                       read(a[z]);
                       if a[z]<=1 then b[z]:=1 else b[z]:=0;
                  end
               else
                   begin
                        built(z*2,x,(x+y) div 2);
                        built(z*2+1,(x+y) div 2+1,y);
                        a[z]:=a[z*2]+a[z*2+1];
                        if (b[z*2]=1) and (b[z*2+1]=1) then b[z]:=1 else b[z]:=0;
                   end;
          end;
function op(z,x,y,l,r:longint):int64;inline;
         var a2,a3,a4:int64;
         begin
              if l>r then exit(0);
              if b[z]=1 then exit(0);
              if (x=l) and (y=r) and (l=r) then
                 begin
                      a2:=a[z];
                      a[z]:=trunc(sqrt(a[z]));
                      if a[z]<=1 then b[z]:=1;
                      exit(a[z]-a2);
                 end;
              a2:=op(z*2,x,(x+y) div 2,l,min((x+y) div 2,r));
              a3:=op(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r);
              a[z]:=a[z]+a2+a3;
              if (b[z*2]=1) and (b[z*2+1]=1) then b[z]:=1;
              exit(a2+a3);
         end;
function cal(z,x,y,l,r:longint):int64;inline;
         begin
              if l>r then exit(0);
              if (x=l) and (y=r) then exit(a[z]);
              exit(cal(z*2,x,(x+y) div 2,l,min(r,(x+y) div 2))+cal(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r));
         end;
procedure swap(var x,y:longint);inline;
          var z:longint;
          begin
               z:=x;x:=y;y:=z;
          end;

begin
     readln(n);
     built(1,1,n);
     readln;
     readln(m);
     for i:=1 to m do
         begin
              readln(j,k,l);
              if k>l then swap(k,l);
              case j of
                   1:writeln(cal(1,1,n,k,l));
                   0:op(1,1,n,k,l);
              end;
         end;
end.